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3 years ago

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Which size bottle of ketchup is the best deal with the lowest unit price? Prices for Different Sizes of Bottles of Ketchup Size

Bottle Price 10 oz. $2.50 14 oz. $2.80 16 oz. $4.00 18 oz. $5.40 a, 10 oz. b, 14 oz. c, 16 oz. d, 18 oz.

1 answer:

butalik [34]3 years ago

5 0

Answer:

a.10 oz

Step-by-step explanation:

2.50/10=10

2.80/14=.20

4.00/16=.25

5.40/18=.30

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Write the inverse variation function given that y varies inversely with x, and y = 0.25 when x = 8.

storchak [24]

Answer:

Step-by-step explanation:

If a given variable varies inversely with another variable, an increase in the value of the given variable would cause a corresponding decrease in the value of the other variable. Also, a decrease in the value of the given variable would cause a corresponding increase in the value of the other variable.

If y varies inversely with x, we would introduce a cost of variation, k so that the expression becomes

y = k/x

When y = 0.25, x = 8

Substituting into the expression above, it becomes

0.25 = k/8

k = 0.25 × 8

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The inverse variation function is

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3 years ago

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

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<span>The standard form of the equation of a circumference is given by the following expression:

</span> $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$
<span>
On the other hand, the general form is given as follows:

</span> $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ <span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span> $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

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<em><u>First. a) matches 5) </u></em>

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

<em><u>Second. b) matches 1) </u></em>

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

<em><u>Third. c) matches 3)</u></em>

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

<em><u>Fourth. d) matches 6)</u></em>

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